Question

Tìm giá trị nhỏ nhất của \(\frac{2010x+2680}{x^2+1}\)

Answer 1

Ta có:\(\frac{2010x+2680}{x^2+1}=\frac{-335x^2-335}{x^2+1}+\frac{335x^2+2010x+3015}{x^2+1}=\)

\(=-335+\frac{335\left(x^2+6x+9\right)}{x^2+1}=-335+\frac{335\left(x+3\right)^2}{x^2+1}\)

Vì \(\left(x+3\right)^2_{min}=0\Leftrightarrow335\left(x+3\right)^2_{min}=0\Leftrightarrow\frac{335\left(x+3\right)^2}{x^2+1}_{min}=0\Leftrightarrow-335+\frac{335\left(x+3\right)^2}{x^2+1}_{min}=-335\)

Vậy \(\frac{2010x+2680}{x^2+1}_{min}=-335\) tại x=-3.